and noting that \(\partial _x |\mathcal {X}_1|^2=-\partial _t |\mathcal {X}_2|^2\) ,
one can prove that the Lax pair () is gauge equivalent to [1]}
\(& \tilde{\Phi }_{x}=M\tilde{\Phi },~~M=\left(\begin{array}{cc}-\frac{i}{2} \lambda ^2 & i\lambda q \\i\lambda q^* &\frac{i}{2} \lambda ^2\end{array}\right),\\& \tilde{\Phi }_{t}=N \tilde{\Phi },~~N= \left(\begin{array}{cc}i(|\mathcal {X}_1|^2-\frac{\lambda ^{-2}}{2}) & i\lambda ^{-1}\mathcal {X}^*_1e^{-i\beta }\\i\lambda ^{-1}\mathcal {X}_1e^{i\beta } & -i(|\mathcal {X}_1|^2-\frac{\lambda ^{-2}}{2})\end{array}\right),\) 
